 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Four From Arithmetic and Geometric (Posted on 2016-04-29) Each of X, Y and Z is a distinct positive integer such that
X, Y and Z are in arithmetic sequence, and
X, Y and Z+2016 are in geometric sequence.

Find the four smallest values of X+Y+Z.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) More infinite solutions Comment 3 of 3 | And more solutions

Starting with Jer's y=x(+/-)x*sqrt(2016/x)

The (-) gives positive solutions if x = 2016*n^2/m^2  where m = 1,2,4,3,6 or 12 (because sqrt(2016) = 12*sqrt(14)
Let z = n/m
Then x = 2016(z^2)
y = 2016(z^2 - z)
z = 2016(z^2 - 2z) = 2016z(z-2)

In order for z to be positive, z must be > 2
The smallest z that therefore works is 25/12

z = 25/12 yields 2016*(625/144, 325/144, 25/144) = (8750, 4550, 350).
The Geometric progression is 8750,4550,2366, with a ratio of 13/25  (a.k.a., 1 - 1/z)

Total x + y + z = 13650, so this does not beat any entries on Jer's list, the last entry of which is 12096

 Posted by Steve Herman on 2016-04-30 11:25:23 Please log in:

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